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# relativity problem

Vince S. writes:

Then there are two ways to get the right answer:

Method A: simple physics in the earth frame
Method B: the Lorentz transformations.
There seems to be at least one more way to get the correct answer:

Method C: Treat the problem as a case of the relativistic Doppler shift.

In this method consider the "light" signal from the tail of the spacecraft
as a (outgoing spherical) radio wave whose wavelength is 200 m (the rocket's
length) in the rocket's rest frame. Calculate the period of this wave; it is
the time in the answer to part a of the problem. Now consider the period of
this wave as viewed from an Earthbound observer watching the rocket recede
from her. This period is lengthened (red-shifted) by a Doppler factor of
sqrt((1+0.95)/(1-0.95)) = sqrt(39) = 6.245. (Note that
4.166X10^-6 s = 6.245 x 6.671x10^-7 s). The red-shift formula rather than
the blue-shift formula is used because for this problem we assume that the
earthbound observer is located at the rocket tail (source) when the initial
wave-crest is emitted and that source is moving away from the earthbound
observer as the wave cycle is being emitted. The tail is to be coincident
with the earthbound observer at the emission of the initial wave-crest since
the time interval measured in both frames should be measured with coincidently
reset clocks.

David Bowman
Georgetown College
dbowman@gtc.georgetown.ky.us